The Jan 1, 2012 issue of The College Mathematics Journal featured articles inspired by the late-great Martin Gardner. One by Ian Stewart begins with the Three Penny Puzzle invented by Gardner and Karl Fulves:
Gardner’s three-penny trick:
The trick is performed by a blindfolded magician. A volunteer places three pennies
in a row, and chooses at will whether each coin shows heads or tails. However, both heads and tails must appear, otherwise the trick ends before it begins. The magician announces that even though she cannot see the coins, she will give instructions to turn coins over so that all three coins show the same face, heads or tails.
The instructions are:
1. Flip the left-hand coin.
2. Flip the middle coin.
3. Flip the left-hand coin.
After steps 1 and 2 the magician asks whether all three coins show the same face,
and if the answer is ‘yes’, the trick stops, otherwise the magician requests the third flip.
Although it is plausible that enough flips will eventually
get all coins the same way up, it is a little surprising that at most three flips are needed.
Not being very good at card tricks and such, I would sometimes perform this trick in study hall or odd class moments to amuse my students, and then challenge them to duplicate the trick. I avoided doing it more than once in order not to have them catch on to the fact that the same pattern always works (why children). In all the many times I did it, and sometimes had students discover the logic, I don't think I ever asked them if they could do the same with four pennies, or five, or n. That is, for n Pennies, what is the minimal number of blind-flips f(n) to get them all alike, and what is the sequence of flips.
I think if three pennies can always be solved with three flips, then four -pennies can be solved with four flips flips at most assuming we keep the same rule that the first three coins are not all the same (probably best to disguise this as no three coins in a row are the same). I also think you need to be able to ask, "Are the first three the same, and also if the first four are the same, which really gives the ending away.
My method that I think will work is as follows:
Since the first three can not be all the same, begin as if there were only three coins. After each flip ask if the first three are the same. If not, continue as if solving the three coin problem, and after the third flip just ask if all four are the same, and flip the fourth coin, if needed. Switching from the first three to the last three and manipulating the three coin solution from left to right, or right to left somewhat at randomly will help obscure the obvious.
My non-proven method for five is to begin with the left three in regular style, asking each time, "are the first three the same?" When they are, switch to playing from the right end and this forces the last two to match the third, and thus all five alike.
Example starting HTHHT , proceed flip coin 1; TTHHT, Flip coin two THHHT; Flip Coin 1, HHHHT
Now you have first three alike, and working from the right end, We flip the first on the right and we have five alike.
If the start had been HTHTT , TTHTT, THHTT, HHHTT, HHHTH, HHHHH. So far it works for all the things I've tried, but I have not systematically exhausted all possible legal starting positions.
I have had no success with even numbers of coins and am still experimenting. If anyone has a solution they can share I would love to receive it. I think the above is extendable in a similar approach for 2n+1 numbers of coins but have not had loads of time to experiment.....yet.
Share your ideas, please.
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